![SearchIcon](https://vmkt.vedantu.com/vmkt/PROD/png/bdcdbbd8-08a7-4688-98e6-4aa54e5e0800-1733305962725-4102606384256179.png)
A lift of mass 1000 Kg which is moving with acceleration 1 m/$s^2$ in upward direction, then the tension developed in string which is connected to lift is:
$\left( A \right)$ 9,800 N
$\left( B \right)$ 10,800 N
$\left( C \right)$ 11,000 N
$\left( D \right)$ 10,000 N
Answer
502.2k+ views
- Hint: In this question, use the property that the net acceleration working on the lift is the sum of the acceleration of the lift and the acceleration due to gravity so use this concept to reach the solution of the question.
Complete step-by-step solution -
According to Newton's second law of motion the force acting on the object is the product of object mass and the net acceleration working on the object.
$ \Rightarrow F = m\left( {{a_{net}}} \right)$
As the lift goes upward (i.e. it is working against the Earth’s gravity) so the total acceleration acting on the lift is the sum of Earth gravity and the acceleration of the lift by which it is going upwards.
Now as we know earth gravity g = 9.8 m/$s^2$.
And it is given that the lift is moving upward with an acceleration of 1 m/$s^2$.
So the net acceleration working on the lift is \[\left( {9.8 + 1} \right) = 10.8\] m/$s^2$.
And it is also given that the mass of the lift is 1000 kg
So the force acting on the lift is
$F = 1000 \times 10.8 = 10800$N
Now the force acting on the lift is the same as tension developed in the string which is connected to the lift as these forces are equal in nature whenever any object is lifted by the cable or string. So the tension working on the string is the same as the force exerted on the object.
So the tension acting on the string = 10,800 N.
So this is the required answer.
Hence option (B) is the correct answer.
Note: Whenever we face such types of question the key concept we have to remember is that the net acceleration acting on the body is the sum of acceleration of the object and the and the acceleration due to gravity i.e. Earth’s gravity when the object is moving upwards and when the object is moving downwards the net acceleration acting on the body is the difference of acceleration due to gravity and the and the acceleration of the object.
Complete step-by-step solution -
According to Newton's second law of motion the force acting on the object is the product of object mass and the net acceleration working on the object.
$ \Rightarrow F = m\left( {{a_{net}}} \right)$
As the lift goes upward (i.e. it is working against the Earth’s gravity) so the total acceleration acting on the lift is the sum of Earth gravity and the acceleration of the lift by which it is going upwards.
Now as we know earth gravity g = 9.8 m/$s^2$.
And it is given that the lift is moving upward with an acceleration of 1 m/$s^2$.
So the net acceleration working on the lift is \[\left( {9.8 + 1} \right) = 10.8\] m/$s^2$.
And it is also given that the mass of the lift is 1000 kg
So the force acting on the lift is
$F = 1000 \times 10.8 = 10800$N
Now the force acting on the lift is the same as tension developed in the string which is connected to the lift as these forces are equal in nature whenever any object is lifted by the cable or string. So the tension working on the string is the same as the force exerted on the object.
So the tension acting on the string = 10,800 N.
So this is the required answer.
Hence option (B) is the correct answer.
Note: Whenever we face such types of question the key concept we have to remember is that the net acceleration acting on the body is the sum of acceleration of the object and the and the acceleration due to gravity i.e. Earth’s gravity when the object is moving upwards and when the object is moving downwards the net acceleration acting on the body is the difference of acceleration due to gravity and the and the acceleration of the object.
Recently Updated Pages
Master Class 11 Economics: Engaging Questions & Answers for Success
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Master Class 11 Business Studies: Engaging Questions & Answers for Success
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Master Class 11 Accountancy: Engaging Questions & Answers for Success
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
The correct geometry and hybridization for XeF4 are class 11 chemistry CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Water softening by Clarks process uses ACalcium bicarbonate class 11 chemistry CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
With reference to graphite and diamond which of the class 11 chemistry CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Trending doubts
10 examples of friction in our daily life
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
One Metric ton is equal to kg A 10000 B 1000 C 100 class 11 physics CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Difference Between Prokaryotic Cells and Eukaryotic Cells
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
State and prove Bernoullis theorem class 11 physics CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
What organs are located on the left side of your body class 11 biology CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
How many valence electrons does nitrogen have class 11 chemistry CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)